Section 6.1 Exercise 10: Find a unit vector u in the direction of the given vector −6 w = 4 . −3 Solution. There are two solutions: √ −6 −6/ 61 ±1 ±1 √ u = w = √ 4 = ± 4/ 61 . kwk 36 + 16 + 9 √ −3 −3/ 61 Exercise 24: Verify the paralellogram law ku + vk2 + ku − vk2 = 2kuk2 + 2kvk2. Solution. Let u = (u1, . , un) and v = (v1, . , vn). We have n n n 2 2 X 2 X 2 X 2 2 2 2 ku + vk + ku − vk = (ui + vi) + (ui − vi) = (2ui + 2vi ) = 2kuk + 2kvk . i=1 i=1 i=1 (Here we used the simple identity (a + b)2 + (a − b)2 = 2a2 + 2b2 valid for any scalars a, b.) Section 6.2 Exercise 30: Let U be orthonormal matrix, and construct V by interchanging some of the rows of U. Explain why V is orthonormal. Solution 1. Let us recall some implications proved in the class. By definition, a matrix U is orthogonal if and only if U T U = I. Since both U and U T are square matrices, the latter identity is equivalent to UU T = I by the Invertible Matrix Theorem. But UU T = I is equivalent to the rows of U being orthonormal. In summary, U is othogonal if and only if its rows are orthonormal. The latter property is clearly preserved by any row permutation. Solution 2 (or rather a hint). Observe that the dot product x · y does not change if the entries of x are permuted in the same way as the entries of y. Sample Solutions Section 6.3 Exercise 7: Let W = Span{u2, u2}. Write y as the sum of a vector in W and a vector orthogonal to W . Here 1 1 5 y = 3 , u1 = 3 , u2 = 1 . 5 −2 4 Solution. The vectors u1 and u2 are orthogonal to each other. First we compute yˆ, the orthogonal projection of y onto W : 10/3 1 + 9 − 10 5 + 3 + 20 2 yˆ = u1 + u2 = 0 u1 + u2 = 2/3 . 1 + 9 + 4 25 + 1 + 16 3 8/3 Let v = y − yˆ = (−7/3, 7/3, 7/3). Then y = yˆ + v is the required sum. Exercise 8: Let W = Span{u2, u2}. Write y as the sum of a vector in W and a vector orthogonal to W . Here −1 1 −1 y = 4 , u1 = 1 , u2 = 3 . 3 1 −2 Solution. The vectors u1 and u2 are orthogonal to each other. First we compute yˆ, the orthogonal projection of y onto W : 3/2 −1 + 4 + 3 1 + 12 − 6 1 yˆ = u1 + u2 = 2 u1 + u2 = 7/2 . 1 + 1 + 1 1 + 9 + 4 2 1 Let v = y − yˆ = (−5/2, 1/2, 2). Then y = yˆ + v is the required sum. Section 6.4 Exercise 10: Find an orthogonal basis for the column space of −1 6 6 3 −8 3 A = [x1 x2 x3] = . 1 −2 6 1 −4 −3 Solution. We apply the Gram-Schmidt process. We let v1 = x1. Also, 3 −6 − 24 − 2 − 4 1 v2 = x2 − v1 = x2 + 3v1 = . 1 + 9 + 1 + 1 1 −1 Finally, we should let −1 −6 + 9 + 6 − 3 18 + 3 + 6 + 3 1 5 −1 v3 = x3 − v1 − v2 = x3 − v1 − v2 = . 1 + 9 + 1 + 1 9 + 1 + 1 + 1 2 2 3 −1 A routine checking shows that the obtained vectors v1, v2, v3 are indeed orthogonal. Section 6.5 Exercise 12: Find (a) the orthogonal projection of b into ColA and (b) a least-square solution of Ax = b. Here 1 1 0 2 1 0 −1 5 A = [v1 v2 v3] = , b = . 0 1 1 6 −1 1 −1 6 Solution. It is easy to check that the columns of A are orthogonal to each other. (In particular, they are linearly independent.) Hence, we can use the standard formulas for finding the orthogonal projection of b onto ColA: 5 ˆ 2 + 5 − 6 2 + 6 + 6 −5 + 6 − 6 1 2 b = v1 + v2 + v1 = (v1 + 14v2 − 5v3) = . 1 + 1 + 1 1 + 1 + 1 1 + 1 + 1 3 3 6 This answers (a). Since the columns of A are linearly independent, the least-square solution x is unique and we already know the weights, namely x(= 1/3, 14/3, −5/3). ˆ As an check, one can compute b − b = (−3, 3, 3, 0) and see that it is indeed orthogonal to each vi..